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3x(2x+10)=54
We move all terms to the left:
3x(2x+10)-(54)=0
We multiply parentheses
6x^2+30x-54=0
a = 6; b = 30; c = -54;
Δ = b2-4ac
Δ = 302-4·6·(-54)
Δ = 2196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2196}=\sqrt{36*61}=\sqrt{36}*\sqrt{61}=6\sqrt{61}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-6\sqrt{61}}{2*6}=\frac{-30-6\sqrt{61}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+6\sqrt{61}}{2*6}=\frac{-30+6\sqrt{61}}{12} $
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